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We prove both the general modular intersection bound for a set family and its sharper uniform version, the Frankl-Wilson theorem. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a family of subsets of $[n]$. Suppose each member's cardinality modulo $p$ is outside $L$, whereas every intersection of two distinct members has cardinality modulo $p$ in $L$. Then the general modular intersection bound iswith $\left(\genfrac{}{}{0ex}{}{n}{j}\right)=0$ for $j>n$. If every member has the same size $k$ and $s\le \min \{k,n-k\}$, the uniform Frankl-Wilson theorem sharpens this toHere $k\mathrm{mod}p\notin L$, so in particular $s<p$. Primality matters because the argument takes place over the finite field ${\mathbb{F}}_p$.

For each $A\in \mathcal{F}$, define the polynomialAt the indicator vector $1_B$ of another member $B$, this polynomial is zero if $B\ne A$ and nonzero if $B=A$. Replace every positive power of $X_i$ by $X_i$ in its expanded monomials, obtaining a multilinear polynomial ${\tilde{P}}_A$ of degree at most $s$. This multilinear reduction on the Boolean cube preserves values at every zero-one vector.

The resulting polynomials are linearly independent. Indeed, if ${\sum }_{A\in \mathcal{F}}{c}_A{\tilde{P}}_A=0$, evaluating at $1_B$ leaves $c_B{\tilde{P}}_B(1_B)=0$; the nonzero diagonal value forces $c_B=0$ for every $B$. But multilinear polynomials of degree at most $s$ lie in the vector space with basis $\{{\prod }_{i\in S}{X}_i:|S|\le s\}$, whose dimension is ${\sum }_{j=0}^s\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The bound on the dimension of a vector space gives the general modular intersection bound for a set family. This is the polynomial method in combinatorics: the intersection conditions produce an evaluation matrix with nonzero diagonal and zero off-diagonal entries.

To prove the sharper uniform bound, we first establish injectivity of inclusion between adjacent set layers. Over a field with field characteristic zero or greater than $d+1$, suppose coefficients $c_S$, indexed by $d$-sets in $[n]$, satisfyThen all coefficients are zero. For $d=0$ this is immediate. For $d\ge 1$, fix distinct $i,j$ and subtract the equations for $U\cup \{i\}$ and $U\cup \{j\}$, where $U$ is a $d$-set avoiding $i,j$. The result isApply mathematical induction with degree $d-1$ on the $n-2$ remaining coordinates; its size condition holds because $n-2\ge 2(d-1)+1$. This forces $c_{S\cup \{i\}}=c_{S\cup \{j\}}$ for every such $S$. Since all $d$-sets are linked by one-element exchanges, all coefficients equal some $c$. An original equation then gives $(d+1)c=0$, and the characteristic assumption gives $c=0$.

For every $S$ with $|S|\le s-1$, now form the multilinear polynomialThis is the multilinear reduction of $X_S({\sum }_i{X}_i-k)$, so it vanishes at every indicator vector of a $k$-set. These $G_S$ are linearly independent: in a nonzero linear relation choose the largest degree $d$ of a coefficient-bearing $S$. The degree-$(d+1)$ terms have coefficients ${\sum }_{S\subset T,|S|=d}{c}_S$. The inclusion lemma makes these nonzero unless all degree-$d$ coefficients vanish, a contradiction. It applies because $d\le s-1$, $p>s$, and $n\ge 2s$.

The $G_S$ are also independent jointly with the ${\tilde{P}}_A$. Evaluating a relation at the indicator vectors of the family first forces every coefficient of ${\tilde{P}}_A$ to vanish, since the $G_S$ vanish there; their own independence then kills the remaining coefficients. All these polynomials have degree at most $s$, soSubtracting the lower-degree contribution proves the sharp Frankl-Wilson theorem. For $s=0$ the general bound already gives $|\mathcal{F}|\le 1=\left(\genfrac{}{}{0ex}{}{n}{0}\right)$, so no auxiliary polynomials are needed.

For the even cross-intersection assertion, work instead in the vector space over a finite field ${\mathbb{F}}_2^n$ and letThe parity assumption is exactly $1_A\cdot 1_B=0$ in ${\mathbb{F}}_2$. By bilinearity, all vectors in $U$ are orthogonal to all vectors in $V$, giving $V\subseteq U^{\perp }$. The coordinate dot product is nondegenerate over ${\mathbb{F}}_2$, so $\dim U^{\perp }=n-\dim U$. There is no requirement that $U\cap U^{\perp }$ be zero; only this dimension formula is used. Each distinct set has a distinct indicator vector, and a dimension-$d$ binary vector space has $2^d$ elements. Therefore the even cross-intersection bound isThis is the even cross-intersection bound, including empty families, for which the product is zero.

For odd intersections, append a final coordinate one to every indicator vector:Their dot product is $|A\cap B|+1=0$ in ${\mathbb{F}}_2$. Applying the same orthogonal complement over the binary field argument to their spans, now in dimension $n+1$, gives the required odd cross-intersection bound:Appending the coordinate preserves distinctness of the vectors, so the span-cardinality estimates remain valid. This proves the odd cross-intersection bound directly by reducing the odd condition to an even one in a larger vector space.

Write $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative and $q=\mathbf{u}\cdot \mathbf{B}$ for the cross-helicity density, so $H_c={\int }_{V}qdV$. The ideal magnetohydrodynamic induction equation of ideal magnetohydrodynamics, with $\nabla \cdot \mathbf{B}=0$, givesThe dot product of the Euler momentum equation with the magnetic field has no Lorentz force contribution, because $\mathbf{B}\cdot [(\nabla \times \mathbf{B})\times \mathbf{B}]=0$. Consequently,Here $h=e+p/\rho$ is the specific enthalpy, $T$ the temperature, and $s$ the specific entropy; $e$ is the internal energy per unit mass. The first law of thermodynamics in this convention is $de=Tds-pd(1/\rho )$, so $dh=Tds+dp/\rho$. Substitution, followed by $\nabla \cdot \mathbf{B}=0$, yields the cross-helicity conservation lawThe vector triple product identity $\mathbf{u}\times (\mathbf{u}\times \mathbf{B})=\mathbf{u}q-u^2\mathbf{B}$ puts this in the equivalent form

For the fixed volume, the divergence theorem gives the exact cross-helicity balanceThus cross-helicity is constant when the net source equals the outward flux. Simple sufficient hypotheses are $\mathbf{B}\cdot \nabla s=0$ throughout the volume and $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ on its boundary. The first condition includes a homentropic flow; the second prevents both relevant boundary fluxes. Periodic boundary conditions, or sufficiently rapid decay at infinity, provide alternatives. These statements concern smooth ideal magnetohydrodynamics: material specific entropy conservation alone, $Ds/Dt=0$, does not eliminate spatial specific entropy gradients along the magnetic field lines.

Let the constant mass density be $\rho >0$ and put ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential $\Phi$ remains in $\psi$; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$; each Elsässer variable is transported by the other.

Define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product with $2{\mathbf{z}}^{\pm }$ gives the Elsässer energy invariant balanceThe divergence theorem proves $d{I}_{\pm }/dt=0$ whenever the net boundary flux vanishes. In particular, $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ makes ${\mathbf{z}}^{\pm }\cdot \mathbf{n}=0$, so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.

The kinetic energy plus magnetic energy isExpanding $|\mathbf{u}\pm {\mathbf{v}}_a{|}^2$ and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity $E$ here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.

Let $\mathbf{n},\mathbf{w},\mathbf{u}$ be orthonormal unit vectors north, west and upward. A direction of altitude $\gamma$ and westward azimuth $\eta$ isThe north celestial pole and the upper equatorial meridian direction areIn the equatorial coordinate system, the same unit vector is $\mathbf{x}=\sin \delta \mathbf{p}+\cos \delta (\cos h\mathbf{e}+\sin h\mathbf{w})$. Taking dot products with $\mathbf{p},\mathbf{w},\mathbf{e}$ respectively givesThis is an orthogonal transformation between two bases, so it preserves the length of the direction vector. Recover $h$ from both its sine and cosine to keep the correct quadrant. At the zenith, azimuth itself is undefined; the vector equations remain meaningful by continuity. Using the more usual eastward azimuth changes the sign of the second equation.

The dot product and Euclidean norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_i{x}_i{y}_i$ and $|\mathbf{x}|=\sqrt{\mathbf{x}\cdot \mathbf{x}}$. For every real $t$,Its quadratic discriminant is nonpositive, proving the Cauchy-Schwarz inequality $|\mathbf{x}\cdot \mathbf{y}|\le |\mathbf{x}||\mathbf{y}|$. In the inequality asked for, equality holds exactly when $\mathbf{x}$ is a positive scalar multiple of $\mathbf{y}$, and

Using the Levi-Civita symbol and its contraction identity,

Put $T=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$, the scalar triple product. Cyclic symmetry gives $\mathbf{m}\cdot \mathbf{a}=\mathbf{m}\cdot \mathbf{b}=\mathbf{m}\cdot \mathbf{c}=T$, so each requested angle $\varphi$ satisfies $\cos \varphi =T/|\mathbf{m}|$. If every pairwise angle is $\theta$ and $t=\cos \theta$, the preceding identity givesand therefore $|\mathbf{m}|=\sqrt{3}(1-\cos \theta )$. This is $0$ when $\theta =0$; for a right-handed orthonormal basis and $\theta =\pi /2$, $\mathbf{m}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ has norm $\sqrt{3}$.

The three planes have equations $\mathbf{a}\cdot \mathbf{x}=\mathbf{a}\cdot \mathbf{p}$, $\mathbf{b}\cdot \mathbf{x}=\mathbf{b}\cdot \mathbf{q}$, and $\mathbf{c}\cdot \mathbf{x}=\mathbf{c}\cdot \mathbf{r}$. They meet at one point exactly when $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$, meaning their normals are linearly independent vectors. The point is

The repulsive force is $\mathbf{F}=k\mathbf{r}/r^3=-\nabla (k/r)$, so the Hamiltonian isThe Hamilton's equations give $\dot{\mathbf{r}}=\mathbf{p}/m$ and $\dot{\mathbf{p}}=k\mathbf{r}/r^3$. Consequentlyso the angular momentum is conserved. Moreover,It follows that the Laplace-Runge-Lenz vector $\mathbf{A}=\mathbf{p}\times \mathbf{L}+mk\hat{\mathbf{r}}$ is conserved. By the Hamiltonian conservation law, these time derivatives are exactly $\{\mathbf{L},H\}$ and $\{\mathbf{A},H\}$, so both Poisson brackets vanish.

The integrals of motion are the energy $H$, the components of $\mathbf{L}$, and the components of $\mathbf{A}$, subject toTaking the dot product of $\mathbf{A}$ with $\hat{\mathbf{r}}$ givesTherefore the polar equation of a repulsive inverse-square orbit iswhere $\lambda =L^2/(mk)\ge 0$ and $e=A/(mk)\ge 0$. For a nonradial physical scattering orbit $H>0$, the identity above gives $e>1$, as expected for a hyperbolic Kepler orbit.

For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, Newton's law of universal gravitation gives the two-body problem in its reduced one-body form,Taking the dot product with the relative velocity $\mathbf{v}=\dot{\mathbf{r}}$ givesIntegration therefore yields conservation of specific orbital energy:

For either sign, the dot product with $v_1$ isThe remaining pair also satisfiesHence the three eigenvectors are mutually orthogonal. This also follows from the spectral theorem for real symmetric matrices, since $M=M^T$ and its eigenvalues are distinct, but the displayed products verify the actual chosen vectors directly.

Put $d=b\cdot x$ and $B=|b{|}^2$. Taking the dot product of the original equation with $b$ eliminates $b\times x$ and gives $d=a\cdot b-Bd$. ThusNext take its cross product with $b$. The term $db$ disappears, and the vector triple product identity givesInsert this expression back into $x=a-db-b\times x$. The two longitudinal terms cancel, leaving $(1+B)x=a-b\times a$. ConsequentlyThe denominator is strictly positive for every real $b$, including $b=0$. Substitution using the same triple-product identity verifies the solution and establishes uniqueness. This is the cross-product operator with a longitudinal rank-one term.

Let $d=b\cdot x$. Taking a dot product with $b$ reduces the vector equation to $(1-b\cdot c)d=b\cdot a$. If $b\cdot c\ne 1$, this determines a unique scalar and hence a unique vector:Direct substitution verifies it, so necessity has also produced sufficiency. The formula includes $c=0$, when $x=a$.

If $b\cdot c=1$, the reduced scalar equation becomes $0=b\cdot a$. When $b\cdot a\ne 0$, there is no solution. When $b\cdot a=0$, all solutions are $x=a+dc$ with arbitrary $d\in \mathbb{R}$: their dot product with $b$ is indeed $d$. Since $b\cdot c=1$ forces $c\ne 0$, this is an entire affine line, not a single point. These alternatives are the solvability of an affine rank-one vector equation.

Differentiate the sum of kinetic energy and potential energy using the dot product product rule:The equations of motion make the bracket zero. Hence the mechanical energy is constant in time. The central restoring force can exchange kinetic and potential energy but supplies no change in their sum.

The three columns of $N=\sqrt{8}A$ are $(\sqrt{3},1,2)$, $(-\sqrt{2},\sqrt{6},0)$, and $(-\sqrt{3},-1,2)$. Each has squared Euclidean norm eight. Their pairwise dot products areHence $N^{\mathsf{T}}N=8I$, so $\boxed{A^{\mathsf{T}}A=I}$. Thus $A$ is an orthogonal matrix, with $A^{-1}=A^{\mathsf{T}}$. The determinant from (i) places it in the determinant-one component.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=(-1,1,0)$. For the first field, ${\mathbf{F}}_1=(-t,1-t,-1+t)$ and its dot product with the tangent is one. ThereforeFor the second field, use the fundamental theorem for line integrals. The endpoints both have $r=1$, so their potential values agree andDirectly, its parameter integrand is $(2t-1)/[1+(1-t{)}^2+t^2{]}^{3/2}$, which changes sign under $t\mapsto 1-t$ and integrates to zero.

The shorter circular arc lies in the $x$-$y$ plane and can be written $\mathbf{r}(\theta )=(\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. The first field is $(-\sin \theta ,\cos \theta ,-\cos \theta )$, whose dot product with ${\mathbf{r}}^{\prime }$ is one. HenceThe second field is radial while the circular tangent is perpendicular to the radius, so $\boxed{{\int }_{\mathrm{arc}}{\mathbf{F}}_2\cdot d\mathbf{r}=0}$. Thus the first field gives different values, $1$ and $\pi /2$, along the two paths, whereas the conservative second field gives the same endpoint difference.

As a check using Stokes theorem, the arc followed by the reversed segment encloses area $\pi /4-1/2$ in the plane. The constant curl has upward component two, giving circulation $2(\pi /4-1/2)=\pi /2-1$, exactly the difference above.

The dot product is $\mathbf{a}\cdot \mathbf{b}=-1+2+3=4$, while the Euclidean norms are $\sqrt{3}$ and $\sqrt{14}$. Thus the angle in the stated range isThe positive cosine shows that it is an acute angle.

The chosen eigenvectors have dot productThey are therefore orthogonal, and their unit lengths make them an orthonormal eigenbasis. This agrees with the spectral theorem for real symmetric matrices, since $A$ is real symmetric and its two eigenvalues are distinct.

Hold $r$ fixed and differentiate the position vector $\mathbf{X}(\theta ,\varphi )$. Its coordinate tangent vectors areTheir dot product is zero, and their squared Euclidean norms are $r^2$ and $r^2{\sin }^2\theta$. To first order, the displacement is $\delta \mathbf{X}={\mathbf{X}}_{\theta }\delta \theta +{\mathbf{X}}_{\varphi }\delta \varphi +O((|\delta \theta |+|\delta \varphi |{)}^2)$. Squaring givesThis is the line element on a sphere. The chord distance and the short surface arc length agree to the order retained here.

For the particular eigenvectors chosen in (iii), the dot products areThus only ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are orthogonal for this choice. The matrix is not symmetric, so distinct eigenvalues alone do not imply orthogonality. Another basis within the repeated eigenspace may change which chosen vectors are perpendicular; the dot products above specify the convention completely.

Perpendicular vectors have zero dot product. Here $\mathbf{u}\cdot \mathbf{v}=3\cdot 4-2c$, so $12-2c=0$. Therefore